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Homework Help: Trig questions

  1. Jan 24, 2006 #1
    1. Express [tex] \frac{sin^2 2 \theta}{1+cos^2 2 \theta} [/tex] as a function of [tex]sin \theta[/tex]

    here's what I did:
    [tex] = \frac{4 sin^2 \theta cos^2 \theta}{1+(2cos^2 \theta -1)^2} [/tex]
    [tex] = \frac{4 sin^2 \theta cos^2 \theta}{1+(4cos^4 \theta - 4 cos^2 \theta + 1)} [/tex]
    [tex] = \frac{2sin^2 \theta cos^2 \theta}{2cos^2 \theta - 2cos^2 \theta +1} [/tex]

    is this correct? can I simplify it more?

    2. Write [tex]sin(a+b)sin(a-b)[/tex] as a function of double angles

    I used the sum and product formulas to simplify the equation but I did not use the double angle formulas. I'm not quite sure what the question is asking.

    Here's what I did:
    [tex] = .5 [cos(a+b-a+b)-cos(a+b+a-b)] [/tex]
    [tex] = .5 [cos(2b)-cos(2a)] [/tex]

    not sure where the double anges come in. any ideas?

    3. simplify [tex] \sqrt{2-2cos4 \Theta}[/tex]

    i can make it become 1-cos4x but I dont know how to simplify it further because of the cos4. no clue on this one, any help would be appreciated.
    Last edited: Jan 24, 2006
  2. jcsd
  3. Jan 24, 2006 #2


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    Homework Helper

    Well it's not written as a function of only sin(x), so it's not correct yet.
    Although there may be ways to simplify, you can reduce the sin²2x in function of only sinx like this:

    [tex]\sin ^2 2x = 4\sin ^2 x\cos ^2 x = 4\sin ^2 x\left( {1 - \sin ^2 x} \right)[/tex]

    In the denominator, there's a cos²2x but that's equal to 1-sin²2x, giving you the case above again.

    I'm not sure but perhaps this was the point, you have now rewritten it as a function of the double angles 2a and 2b.

    There are 3 version of the double-angle formule for the cosine, choose the one which makes the constant disappear:

    [tex]\sqrt {2 - 2\cos 4x} = \sqrt {2 - 2\left( {1 - 2\sin ^2 2x} \right)} [/tex]
  4. Jan 24, 2006 #3
    [tex]\frac {\sin^2 2 \theta} {2- \sin^2 2\theta} = \frac 2 {2 - \sin^2 2\theta} - 1[/tex]
    and simplify the denominator using [tex]\sin 2\theta = 2 \sin \theta \cos \theta[/tex] and [tex]\cos^2 \theta = 1 - \sin^2 \theta[/tex]...? Still complicated.

    2. looks correct.

    3. [tex] \sqrt{ 2( 1 - \cos 4\theta)} = \sqrt {2( 1 - \cos^2 2\theta + \sin^2 2\theta)} = 2 | \sin 2\theta} | = 4 | \sin \theta \cos \theta }|[/tex]

    hmmm...I have to go to see a class day of the kindergarten of my daughter next month... :)
    Last edited: Jan 24, 2006
  5. Jan 24, 2006 #4
    Your two problems can be solved easily-------------Akash

    =Sin(a^2) -Sin(b^2) /////////////////////you can check this easily. it is a formula
    =[cos(2a)-cos(2b)]/2 ////////////////expressed in double angle in cos

    (2-2cos4A)^½=(2(1-cos4A))^½ ////////////A IS ANGLE THETA

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